I have ac-bc=0
prove: ac-bc=c(a+(-b)) (I used the axiom associative)
because c ∉ 0 then (a+(-b))=0 so a=b

Can you tell if my pove is right and which axioms I need to use to prove this?
thank you

September 29th 2012, 11:14 AM

HallsofIvy

Re: field axioms

Odd. You titled this "field axioms" but Z is not a field!

Okay, from ac= bc, you get ac- bc= 0 and then you say
"ac-bc=c(a+(-b)) (I used the axiom associative)"
Well, first, that's the distributive axiom, not associative. But also you don't say that is equal to 0 which you should.
Now, which axiom, specifically, allows you to say that "if c(a- b)= 0 and c is not 0 then a- b= 0"?

This part I didn't learn at college, but I will learn it soon.
My teacher said I could use this axiom: voor every a,b in Z with ab=0 it follows that a=0 or b=0
So I see that (a-b)c=0 we say that a-b=0 or c=0 but we now c isn't equal to zero so we can say that a-b=0

Thank you very much for your help!

September 30th 2012, 02:06 PM

Deveno

Re: field axioms

it turns out that there are "kinds of numbers" that go "in a circle", rather than "an infinite line" like the integers do.

because numbers like that exist, it is not a "universal property of rings" (which is the type of algebraic structure the integers ARE) that ab = 0 implies a = 0, or b = 0 (or maybe both).

it is, however, true for POLYNOMIALS with coefficients that are integers (for example). and polynomials with integer coefficients are very much LIKE integers (our arabic numeral system can be thought of as "polynomials in 10 (rather than x)" which explains why polynomial addition and multiplication are so "normal").

i point this out because for algebraic structure that ARE euclidean domains, we can use the very same proof we used above:

in a euclidean domain, if ac - bc = 0, then a = b. the advantage to this is that anything (no matter how "weird" the actual elements of our euclidean domain might be) that satisfies the same rules as the axioms we used in our proof above, also can use this proof. this is why things are often defined using axioms: you don't have to prove things over and over for each new "type of thing (structure)", prove it once as a consequence of the axioms and that takes care of all sorts of different cases at once.

it turns out that integers are fairly pervasive in mathematics. many (but not all) mathematical structures can be seen as "integers interacting with other things". but integers are simple, right? well, actually, no. if all we ever did was add, then yes, integers WOULD be simple. we could just "count everything". but multiplication throws a monkey wrench into the works.

for counting, everything is "1's", we just have different symbols because "5" is easier to write than 1+1+1+1+1. but for multiplying, everything is a combination of "prime numbers", and prime numbers are...unpredictable. there's doesn't seem to be a "pattern" to how often they occur (there is, sort of, but it's very complicated, and we can't prove it, yet).